$$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$ I started off doing $$u =9x$$ to get $$\int\frac{\frac{u}{9}^3}{\sqrt{u^2-16}}dx$$ which allows for the trig substitution of $$x = a\sec\theta$$

making the denomonator $$\sqrt{16\sec^2\theta-16}$$ $$\tan^2\theta + 1 = \sec^2\theta$$ $$\Rightarrow4\tan\theta$$ to give $$\int\frac{\frac{4\sec}{9}^3}{4\tan\theta}dx$$

Am I doing this correctly?


3 Answers 3


If Trigonometric Substitution is not mandatory,

make the following substitution $$\sqrt{81x^2-16}=u$$

$$\implies81x^2-16=u^2\implies162x\ dx=2u\ du\iff x\ dx=\frac{u\ du}{81}$$

and $x^2=\dfrac{u^2+16}{81}$

Hope you can take it from here

  • $\begingroup$ I believe that it is manditory, however, where would you from here? $\endgroup$
    – Jessica
    Sep 28, 2014 at 18:15
  • $\begingroup$ @Jessica, Please find the edited version $\endgroup$ Sep 28, 2014 at 18:17
  • $\begingroup$ Nice answer .... $\endgroup$
    – Semsem
    Sep 28, 2014 at 18:18

Set $9x=4\sec\theta\implies81x^2-16=(4\tan\theta)^2$


$$=\int\frac{4^3\sec^3\theta}{9^3\cdot4\tan\theta\cdot(\text{sign of}\tan\theta)}\frac49\sec\theta\tan\theta\ d\theta$$

$$=\frac{4^3}{9^4\cdot(\text{sign of}\tan\theta)}\int(1+\tan^2\theta)\sec^2\theta\ d\theta$$

Hope you can take it home from here

  • $\begingroup$ I got: $\frac{4^3}{9^3}\bigg(\tan\theta+\frac{\tan^3\theta}{3}\bigg)+C$. Is this correct? $\endgroup$
    – Jessica
    Sep 28, 2014 at 19:01
  • $\begingroup$ @Jessica, Then you need to replace back $x$. Do you find the relation between my two answers? $\endgroup$ Sep 29, 2014 at 4:16
  • $\begingroup$ what would this look like in terms of x? $\endgroup$
    – Jessica
    Sep 29, 2014 at 5:15
  • $\begingroup$ @Jessica, $4\tan\theta=$(sign of $\tan\theta)(\sqrt{81x^2-16})$ $\endgroup$ Sep 29, 2014 at 5:23
  • $\begingroup$ so $\frac{4^3}{9^3}\bigg(\frac{\sqrt{81x^2-16}}{4}+\frac{(\frac{\sqrt{81x^2-16}}{4})^3}{3}\bigg)$? $\endgroup$
    – Jessica
    Sep 29, 2014 at 5:33

First, to answer the question as asked, you forgot about converting from $du$ to $dx$, then from $du$ to $d\theta$. For the first substitution,


So after that step, you should take what you currently had after the first step and divide it by $9$. For the second substitution

$$u=4\sec\theta,du=4\sec\theta\tan\theta d\theta$$

This should leave you with

$$\int\dfrac{\dfrac{4^3\sec^3\theta}{9^4}}{4\tan\theta}(4\sec\theta\tan\theta d\theta)=\int\frac{4^3}{9^4}(\tan^2\theta+1)\sec^2\theta d\theta$$

As an alternate substitution (trigonometric substitution is not always the best way), you could use

$$u=81x^2-16,du=162xdx$$ $$\int\dfrac{x^3dx}{\sqrt{81x^2-16}}=\int\dfrac{\frac1{162}x^2(162xdx)}{\sqrt{81x^2-16}}=\frac1{162}\int\dfrac{\frac{u+16}{81}du}{u^{1/2}}=$$ $$\frac1{13122}\int u^{1/2}du+\frac8{6561}\int u^{-1/2}du$$

  • $\begingroup$ so would this be the answer from what I was working with? $\frac{4^3}{9^3}\bigg(\tan\theta+\frac{\tan^3\theta}{3}\bigg)+C$ $\endgroup$
    – Jessica
    Sep 28, 2014 at 23:32
  • 1
    $\begingroup$ @Jessica Now that I look again, you did the same thing with the first substitution. I'll be updating the answer. $\endgroup$
    – Mike
    Sep 29, 2014 at 5:43
  • $\begingroup$ is this looking good? $$\frac{4^3}{9^3}\bigg[\frac{\sqrt{81x^2-16}}{12}\bigg(3+\frac{81x^2-16}{16}‌ ​\bigg)\bigg]+C$$ $\endgroup$
    – Jessica
    Sep 29, 2014 at 6:16

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